Source: log.go in package math


package math

Floating-point logarithm.

The original C code, the long comment, and the constants below are from FreeBSD's /usr/src/lib/msun/src/e_log.c and came with this notice. The go code is a simpler version of the original C. ==================================================== Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. Developed at SunPro, a Sun Microsystems, Inc. business. Permission to use, copy, modify, and distribute this software is freely granted, provided that this notice is preserved. ==================================================== __ieee754_log(x) Return the logarithm of x Method : 1. Argument Reduction: find k and f such that x = 2**k * (1+f), where sqrt(2)/2 < 1+f < sqrt(2) . 2. Approximation of log(1+f). Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) = 2s + 2/3 s**3 + 2/5 s**5 + ....., = 2s + s*R We use a special Reme algorithm on [0,0.1716] to generate a polynomial of degree 14 to approximate R. The maximum error of this polynomial approximation is bounded by 2**-58.45. In other words, 2 4 6 8 10 12 14 R(z) ~ L1*s +L2*s +L3*s +L4*s +L5*s +L6*s +L7*s (the values of L1 to L7 are listed in the program) and | 2 14 | -58.45 | L1*s +...+L7*s - R(z) | <= 2 | | Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2. In order to guarantee error in log below 1ulp, we compute log by log(1+f) = f - s*(f - R) (if f is not too large) log(1+f) = f - (hfsq - s*(hfsq+R)). (better accuracy) 3. Finally, log(x) = k*Ln2 + log(1+f). = k*Ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*Ln2_lo))) Here Ln2 is split into two floating point number: Ln2_hi + Ln2_lo, where n*Ln2_hi is always exact for |n| < 2000. Special cases: log(x) is NaN with signal if x < 0 (including -INF) ; log(+INF) is +INF; log(0) is -INF with signal; log(NaN) is that NaN with no signal. Accuracy: according to an error analysis, the error is always less than 1 ulp (unit in the last place). Constants: The hexadecimal values are the intended ones for the following constants. The decimal values may be used, provided that the compiler will convert from decimal to binary accurately enough to produce the hexadecimal values shown.

Log returns the natural logarithm of x. Special cases are: Log(+Inf) = +Inf Log(0) = -Inf Log(x < 0) = NaN Log(NaN) = NaN

func Log(x float64) float64

func log(x float64) float64 {
	const (
		Ln2Hi = 6.93147180369123816490e-01 /* 3fe62e42 fee00000 */
		Ln2Lo = 1.90821492927058770002e-10 /* 3dea39ef 35793c76 */
		L1    = 6.666666666666735130e-01   /* 3FE55555 55555593 */
		L2    = 3.999999999940941908e-01   /* 3FD99999 9997FA04 */
		L3    = 2.857142874366239149e-01   /* 3FD24924 94229359 */
		L4    = 2.222219843214978396e-01   /* 3FCC71C5 1D8E78AF */
		L5    = 1.818357216161805012e-01   /* 3FC74664 96CB03DE */
		L6    = 1.531383769920937332e-01   /* 3FC39A09 D078C69F */
		L7    = 1.479819860511658591e-01   /* 3FC2F112 DF3E5244 */
	)

special cases

	switch {
	case IsNaN(x) || IsInf(x, 1):
		return x
	case x < 0:
		return NaN()
	case x == 0:
		return Inf(-1)
	}

reduce

	f1, ki := Frexp(x)
	if f1 < Sqrt2/2 {
		f1 *= 2
		ki--
	}
	f := f1 - 1
	k := float64(ki)

compute

	s := f / (2 + f)
	s2 := s * s
	s4 := s2 * s2
	t1 := s2 * (L1 + s4*(L3+s4*(L5+s4*L7)))
	t2 := s4 * (L2 + s4*(L4+s4*L6))
	R := t1 + t2
	hfsq := 0.5 * f * f
	return k*Ln2Hi - ((hfsq - (s*(hfsq+R) + k*Ln2Lo)) - f)